In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.

Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.

Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.

Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.

How can I get an explicit example of the last assertion?

How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?


Here is an explicit example. Denote by $E$ the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Now let $B=C=E$. The involution that we will consider is $$(z,w)\in (E\times E)\to (z+\frac{1}{2},-w).$$

Notice now that this quotient is not a direct product anymore. You can see this by calculating $H_1(E\times E)/\mathbb Z_2$ and verifying that it is not $\mathbb Z^4$. Alternatively, you can notice that the quotient does not have a non-vanishing holomorphic volume form. Indeed if such form existed it could be lifted to $E\times E$ to a form $cdz\wedge dw$ ($c\ne 0$), which would give a contradiction since $dz\wedge dw$ is anti-invariant.

  • 5
    $\begingroup$ In other words, $ty^2=x^3-x$. $\endgroup$ – Felipe Voloch Feb 20 '10 at 14:16

If $C=B$ with the same $Z/2$ action, and $g(C/Z_2) \geq 2$, then you always get a non-trivial family:

Suppose it was trivial. Then it had another projection $q : D \to C$. Since $C$ is an \'etale cover of $B'$, and both have genus at least two, $g(C) > g(B')$ by the Hurwitz-formula. That is, also by the Hurwitz formula, all the maps from $B'$ to $C$ are the constant maps. So, all sections $B' \to D$ of $f$ would have to be contained in a fiber of $q$, i. e. they have to be one fiber of $q$. However there are two natural sections $E$ and $F$ of $p$: $[c] \to [(c,c)]$ and $[c] \to [(c,-c)]$. So both of these have to be contracted to a point by $q$. Given any $b \in B$, a choice $c \in C$ such that $[c]=b$ gives an isomorphism $a_c : p^{-1}(d) \to C$, which sends $[(c,c')]$ to $c'$. Now one can construct automorphisms $\varphi_{c'}$ of $C$ for any $c' \in C$ by the following composition:

$ C \to^{a_c^{-1}} p^{-1}([c]) \to^q C \to^s p^{-1}([c']) \to^{a_{c'}} C $


$ S=(q|_{p^{-1}([c'])})^{-1} $

One specialty about $\varphi_{c'}$ is that it takes $c$ and $-c$ to $c'$ and $-c'$, respectively. This follows from the fact, that $E$ and $F$ are contracted to a point by $q$. So, $C$ has infinitely many different automorphisms, which is a contradiction by the assumption that $g(C) \geq 2$. That is our assumption is false, therefore $Y$ is not a product family.

  • $\begingroup$ A very nice argument! I have already assigned the right answer mark to Dmitri, who indeed provided a perfectly legitimate example with g=1, but it is very nice to see that the family is never a product for g at least 2. $\endgroup$ – Andrea Ferretti Feb 22 '10 at 0:53
  • $\begingroup$ I did not get your example. The action of $\mathbb{Z}/2$ on $B = C$ is not free. What am I missing? Thanks. $\endgroup$ – F_L Sep 22 '14 at 20:08

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